Question

1. find the absolute maximum and minimum of $f(x)=x^{2}+\frac{2}{x}$ on $\frac{1}{2},2$.

Explanation:

Step1: Find the derivative

$f'(x)=2x - \frac{2}{x^{2}}=\frac{2x^{3}-2}{x^{2}}$

Step2: Set the derivative equal to 0

$\frac{2x^{3}-2}{x^{2}} = 0$, then $2x^{3}-2=0$. Solving for $x$, we have $x^{3}=1$, so $x = 1$.

Step3: Evaluate the function at critical - point and endpoints

• Evaluate $f(x)$ at $x = 1$: $f(1)=1^{2}+\frac{2}{1}=3$.
• Evaluate $f(x)$ at $x=\frac{1}{2}$: $f(\frac{1}{2})=(\frac{1}{2})^{2}+\frac{2}{\frac{1}{2}}=\frac{1}{4}+4=\frac{17}{4}$.
• Evaluate $f(x)$ at $x = 2$: $f(2)=2^{2}+\frac{2}{2}=4 + 1=5$.

Answer:

The absolute minimum is $3$ at $x = 1$, and the absolute maximum is $\frac{17}{4}$ at $x=\frac{1}{2}$.